Đáp án: Giải trực tiếp pt sau -x ² +10 =0 - Đáp án ( S = ) (±√(10)) Giải thích các bước giải -x^2+10=0 ⇔10-x^2=0 ⇔-x^2=-10 ⇔x^2=10 ⇔[ (l)x=√(10)x=-√(10) (Vậy S = ) (±√(10))

Đáp án ( S = ) (±√(10)) Giải thích các bước giải -x^2+10=0 ⇔10-x^2=0 ⇔-x^2=-10 ⇔x^2=10 ⇔[ (l)x=√(10)x=-√(10) (Vậy S = ) (±√(10))

Đáp án + Giải thích các bước giải -x^2 + 10 = 0 <=> -x^2 = -10 <=> x^2 = 10 <=> x = ± √(10) Vậy phương trình có tập nghiệm S = (± √(10))

blackpink12347704

4 tháng trước

Giải trực tiếp pt sau: -x ² +10 =0

Xem đáp án

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baonhivu1234

4 tháng trước

Đáp án:

$\text{ S = }$ $\left\{±\sqrt{10}\right\}$

Giải thích các bước giải:

$-x^2+10=0$

$⇔10-x^2=0$

$⇔-x^2=-10$

$⇔x^2=10$

$⇔\left[ \begin{array}{l}x=\sqrt{10}\\x=-\sqrt{10}\end{array} \right.$

$\text{Vậy S = }$ $\left\{±\sqrt{10}\right\}$

hoangviet2007

4 tháng trước

Đáp án + Giải thích các bước giải:

$-x^2 + 10 = 0$

$<=> -x^2 = -10$

$<=> x^2 = 10$

$<=> x = \pm \sqrt{10}$

Vậy phương trình có tập nghiệm $S = {\pm \sqrt{10}}$

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